A triangle has two corners with angles of $\frac{π}{2}$ $( pi ) / 2$ and $\frac{5 π}{12}$ $( 5 pi )/ 12$ . If one side of the triangle has a length of $9$ $9$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-07-06T09:53:27

The area is $= 151.1 u^{2}$ $=151.1u^2$

The third angle of the triangle is

$= \frac{1}{2} π - \frac{5}{12} π = \frac{1}{12} π$ $=1/2pi-5/12pi=1/12pi$

To have the largest possible area, the side of length $9$ $9$ is opposite
the smallest angle, i.e, $\frac{1}{12} π$ $1/12pi$

Let the side opposite the angle $\frac{5}{12} π$ $5/12pi$ be $= a$ $=a$

Applying the sine rule to the triangle,
$\frac{a}{sin (\frac{5}{12} π)} = \frac{9}{sin (\frac{1}{12} π)}$ $a/sin(5/12pi)=9/sin(1/12pi)$

Therefore,

$a = 9 \cdot \frac{sin (\frac{5}{12} π)}{sin (\frac{1}{12} π)} = 33.6$ $a=9*sin(5/12pi)/sin(1/12pi)=33.6$

The area of the triangle is

$A = \frac{1}{2} \cdot 9 \cdot 33.6 = 151.1$ $A=1/2*9*33.6=151.1$

A triangle has two corners with angles of π2 ( pi ) / 2 and 5π12 ( 5 pi )/ 12 . If one side of the triangle has a length of 99 , what is the largest possible area of the triangle?